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$$\binom{n}{0}+\binom{n}{1}+\binom{n}{2} = 22$$

I'm trying to solve this equation for $n$.

If $$\binom{n}{0} = 1$$

Then we have that

$$1+\binom{n}{1}+\binom{n}{2} = 22 \implies \binom{n}{1}+\binom{n}{2} = 21$$

However, I'm stuck there. Could you help me out?

Regards

Mark
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    Why not write out the binomial symbols? You'll get a polynomial in $n$ you can solve. Or, just use trial and error. – lulu Nov 11 '18 at 13:29
  • @lulu When I used trial and error, I noticed that $n = 6$ satisfies with the condition. – Mark Nov 11 '18 at 13:30
  • So, that's a solution. Shouldn't be too hard to argue that it's the only one. Or, like I say, just write it out as a polynomial equation. It's just a quadratic. – lulu Nov 11 '18 at 13:31

2 Answers2

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Note that$$\binom n0+\binom n1+\binom n2=1+n+\frac{n(n-1)}2.$$Therefore, your equation is a quadratic one.

José Carlos Santos
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Since for $n=7$ we get ${7\choose 2} = 21$ the $n$ can not be $\geq 8$. So you could try to solve it by inspection for all $n\leq 7$.

nonuser
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